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# NCERT Solutions Chapter 4 - Linear Equation In Two Variables (I), Class 9, Maths Class 9 Notes | EduRev

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## Class 9 : NCERT Solutions Chapter 4 - Linear Equation In Two Variables (I), Class 9, Maths Class 9 Notes | EduRev

The document NCERT Solutions Chapter 4 - Linear Equation In Two Variables (I), Class 9, Maths Class 9 Notes | EduRev is a part of Class 9 category.
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1. The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.
(Take the cost of a notebook to be x and that of a pen to be y).

Let the cost of pen be y and the cost of notebook be x.
A/q,
Cost  of a notebook = twice the pen = 2y.
∴2y = x
⇒ x - 2y = 0
This is a linear equation in two variables to represent this statement.

2. Express the following linear equations in the form ax by c = 0 and indicate the values of a, b and c in each case:
(i) 2x + 3y = 9.35

(ii) x - y/5 - 10 = 0

(iii) -2x + 3y = 6

(iv) x = 3y

(v) 2x = -5y

(vi) 3x + 2 = 0

(vii) y - 2 = 0

(viii) 5 = 2x

(i) 2x + 3y = 9.35
⇒ 2x + 3y - 9.35 = 0
On comparing this equation with ax + by + c = 0, we get
a = 2x, b = 3 and c = -9.35

(ii) x - y/5 - 10 = 0
On comparing this equation with ax + by + c = 0, we get
a = 1, b = -1/5 and c = -10

(iii) -2x + 3y = 6
⇒ -2x + 3y - 6 = 0
On comparing this equation with ax + by + c = 0, we get
a = -2, b = 3 and c = -6

(iv) x = 3y
⇒ x - 3y = 0
On comparing this equation with ax + by + c = 0, we get
a = 1, b = -3 and c = 0

(v) 2x = -5y
⇒ 2x + 5y = 0
On comparing this equation with ax + by + c = 0, we get
a = 2, b = 5 and c = 0

(vi) 3x + 2 = 0
⇒ 3x + 0y + 2 = 0
On comparing this equation with ax + by + c = 0, we get
a = 3, b = 0 and c = 2

(vii) y - 2 = 0
⇒ 0x + y - 2 = 0
On comparing this equation with ax + by + c = 0, we get
a = 0, b = 1 and c = -2

(viii) 5 = 2x
⇒ -2x + 0y + 5 = 0
On comparing this equation with ax + by + c = 0, we get
a = -2, b = 0 and c = 5

Exercise 4.2

1. Which one of the following options is true, and why?
y = 3x + 5 has
(i) a unique solution,

(ii) only two solutions,

(iii) infinitely many solutions

Since the equation, y = 3x + 5 is a linear equation in two variables. It will have (iii) infinitely many solutions.

2. Write four solutions for each of the following equations:
(i) 2x + y = 7

(ii) πx + y = 9

(iii) x = 4y

(i) 2x + y = 7
⇒ y = 7 - 2x
→ Put x = 0,
y = 7 - 2 × 0 ⇒ y = 7
(0, 7) is the solution.
→ Now, put x = 1
y = 7 - 2 × 1 ⇒ y = 5
(1, 5) is the solution.
→ Now, put x = 2
y = 7 - 2 × 2 ⇒ y = 3
(2, 3) is the solution.
→ Now, put x = -1
y = 7 - 2 × -1 ⇒ y = 9
(-1, 9) is the solution.
The four solutions of the equation 2x + y = 7 are (0, 7), (1, 5), (2, 3) and (-1, 9).

(ii) πx + y = 9
⇒ y = 9 - πx
→ Put x = 0,
y = 9 - π×0 ⇒ y = 9
(0, 9) is the solution.
→ Now, put x = 1
y = 9 - π×1 ⇒ y = 9-π
(1, 9-π) is the solution.
→ Now, put x = 2
y = 9 - π×2 ⇒ y = 9-2π
(2, 9-2π) is the solution.
→ Now, put x = -1
y = 9 - π× -1 ⇒ y = 9+π
(-1, 9+π) is the solution.
The four solutions of the equation πx + y = 9 are (0, 9), (1, 9-π), (2, 9-2π) and (-1, 9+π).

(iii) x = 4y
→ Put x = 0,
0 = 4y ⇒ y = 0
(0, 0) is the solution.
→ Now, put x = 1
1 = 4y ⇒ y = 1/4
(1, 1/4) is the solution.
→ Now, put x = 4
4 = 4y ⇒ y = 1
(4, 1) is the solution.
→ Now, put x = 8
8 = 4y ⇒ y = 2
(8, 2) is the solution.
The four solutions of the equation πx + y = 9 are (0, 0), (1, 1/4), (4, 1) and (8, 2).

3. Check which of the following are solutions of the equation x - 2y = 4 and which are not:

(i) (0, 2)

(ii) (2, 0)

(iii) (4, 0)

(iv) (√2, 4√2)

(v) (1, 1)

(i) Put x = 0 and y = 2 in the equation x - 2y = 4.
0 - 2×2 = 4
⇒ -4 ≠ 4
∴ (0, 2) is not a solution of the given equation.

(ii) Put x = 2 and y = 0 in the equation x - 2y = 4.
2 - 2×0 = 4
⇒ 2 ≠ 4
∴ (2, 0) is not a solution of the given equation.

(iii) Put x = 4 and y = 0 in the equation x - 2y = 4.
4 - 2×0 = 4
⇒ 2 = 4
∴ (4, 0) is a solution of the given equation.

(iv) Put x = √2 and y = 4√2 in the equation x - 2y = 4.
√2 - 2×4√2 = 4 ⇒ √2 - 8√2 = 4 ⇒ √2(1 - 8) = 4
⇒ -7√2  ≠ 4
∴ (√2, 4√2) is not a solution of the given equation.

(v) Put x = 1 and y = 1 in the equation x - 2y = 4.
1 - 2×1 = 4
⇒ -1 ≠ 4
∴ (1, 1) is not a solution of the given equation.

4. Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.

Given equation = 2x + 3y = k
x = 2, y = 1 is the solution of the given equation.
A/q,
Putting the value of x and y in the equation, we get
2×2 + 3×1 = k
⇒ k = 4 + 3
⇒ k = 7

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